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## How do you Find the Interval where f is Concave Up and Where f is Concave Down for f(x) = – (2x^{3}) – (3x^{2}) – 7x + 2?

We will use the second derivative test to solve this.

## Answer: f(x) is concave up when x < −1/2 and concave down when x > −1/2.

Let's solve this step by step.

**Explanation:**

Given that, f(x) = –(2x^{3}) – (3x^{2}) – 7x + 2

The first derivative of f(x):

d/dx (–2x^{3} – 3x^{2} – 7x + 2) = −2(3)x^{2} −3(2)x −7 = −6x^{2} − 6x − 7

f'(x) = −6x^{2} − 6x − 7

The second derivative of f(x):

d/dx (−6x^{2} − 6x − 7) = −6(2)x − 6 = −12x − 6

Thus, d^{2}f(x) /dx^{2} = f''(x) = −12x − 6

A function is concave up for the intervals where d^{2}f(x) /dx^{2} > 0 and concave down for the intervals where d^{2}f(x) /dx^{2} < 0.

Intervals where f(x) is concave up: −12x − 6 > 0

−12x > 6

⇒ x < −1/2

Intervals where f(x) is concave down: −12x − 6 < 0

−12x < 6

⇒ x > −1/2

### Thus, f(x) is concave up when x < −1/2 and concave down when x > −1/2.

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